Question

How do you differentiate `f(x)=csc(lnx)` using the chain rule?

Answer 1

`(df)/(dx)=-1/xcsc(lnx)cot(lnx)`

Explanation:

As `f(x)=csc(lnx)=csc(g(x))`, where `g(x)=lnx`

As according to chain rule `(df)/(dx)=(df)/(dg)xx(dg)/(dx)`

`(df)/(dx)=-csc(lnx)cot(lnx)xx1/x=-1/xcsc(lnx)cot(lnx)`